$\sum\limits_{n=0}^{\infty }{-\frac{{{( \frac{2}{3} )}^{n}}{{x}^{n}}}{n!}}~~$ is the Maclaurin series for which function? Choose 1 answer: Choose 1 answer: (Choice A) A $-\frac{2}{3}\cos x$ (Choice B) B $\sin\left(\frac{2}{3}x\right)$ (Choice C) C $ -e^{\frac{2}{3}x}$ (Choice D) D $-\frac{2}{3}e^x$ (Choice E) E $ e^{-\frac{2}{3}x}$
Note that the denominators include all possible factorials. This suggests that the given series most closely resembles the Maclaurin series for the function $~{{e}^{x}}$. ${{e}^{x}}=1+x+\frac{{{x}^{2}}}{2!}+...+\frac{{{x}^{n}}}{n!}+...$ However, we have a minus sign in each term, and we have powers of $~\frac{2}{3}x~$ instead of powers of $~x\,$. Thus we see the following. ${-e^{\frac{2}{3}x}}=-1-\frac{2}{3}x-\frac{{\frac{4}{9}{x}^{2}}}{2!}-...-\frac{(\frac{2}{3})^n{{x}^{n}}}{n!}+...~=~\sum\limits_{n=0}^{\infty }{-\frac{{(\frac{2}{3})^{n}}{{x}^{n}}}{n!}}$